科爾多瓦：如何捕捉視頻，然後上傳到PHP服務器（JS + PHP代碼）

Question

網絡或cordova文檔上有例外，但並不總是完整或最新的。

Answer 1

的工作爲例：

在HTML文件中兩個按鈕：

<a data-role="button" onclick="captureVideo()">Capture video</a> 
<a data-role="button" onclick="uploadVideo()">Send me</a> 

在JS文件：

var path = ''; 

function captureVideo() { 
    // capture callback 
    var captureSuccess = function(mediaFiles) { 
     path = mediaFiles[0].fullPath; 
    }; 

    // capture error callback 
    var captureError = function(error) { 
     navigator.notification.alert('Error code: ' + error.code, null, 'Capture Error'); 
    }; 

    // start video capture 
    navigator.device.capture.captureVideo(captureSuccess, captureError, {limit:1}); 
} 

function uploadVideo() { 
    var win = function (r) { 
     console.log("Code = " + r.responseCode); 
     console.log("Response = " + r.response); 
     console.log("Sent = " + r.bytesSent); 
    } 

    var fail = function (error) { 
     alert("An error has occurred: Code = " + error.code); 
     console.log("upload error source " + error.source); 
     console.log("upload error target " + error.target); 
    } 

    var fileURL = path; 

    var options = new FileUploadOptions(); 
    options.fileKey = "file"; 
    options.fileName = fileURL.substr(fileURL.lastIndexOf('/') + 1); 
    options.mimeType = "video/mp4"; 

    var params = {}; 
    params.value1 = "test"; 
    params.value2 = "param"; 

    options.params = params; 

    var ft = new FileTransfer(); 
    ft.upload(fileURL, encodeURI("https://myserver.com/php/upload.php"), win, fail, options); 
} 

的PHP文件upload.php的：

<?php 
header('Access-Control-Allow-Origin: *'); 
header('Access-Control-Allow-Methods: GET, POST'); 

$uploaddir = '../video/'; //folder https://myserver.com/video 
$uploadfile = $uploaddir . basename($_FILES['file']['name']); 

if (move_uploaded_file($_FILES['file']['tmp_name'], $uploadfile)) { 
    echo "File uploaded\n"; 
} 
else { 
    echo "Error\n"; 
} 
?>